Interpret the binary operation

Given $x*y=x^2+y^3$ and $(x*1)*1=x*(1*1).$

Compute both sides

First, $x*1=x^2+1.$ Hence,

Now compute the right-hand side: $1*1=1^2+1^3=2.$ Therefore, $x*(1*1)=x*2=x^2+2^3=x^2+8.$

So the given condition becomes $(x^2+1)^2+1=x^2+8.$

Solve for $x$

Expand: $x^4+2x^2+1+1=x^2+8$ $x^4+x^2-6=0.$

Let $t=x^2$. Then $t^2+t-6=0$ $(t+3)(t-2)=0.$

Since $t=x^2\ge 0$, we get $x^2=2.$

Evaluate the expression inside $\sin^{-1}$

We need $2\sin^{-1}\left(\frac{x^4+x^2-2}{x^4+x^2+2}\right).$

Using $x^2=2$, we have $x^4=4$. Thus \frac{x^4+x^2-2}{x^4+x^2+2}=rac{4+2-2}{4+2+2}=rac{4}{8}=rac12.

Hence the expression becomes $2\sin^{-1}\left(\frac12\right).$

Use principal value of inverse sine

$\sin^{-1}\left(\frac12\right)=\frac{\pi}{6}.$ Therefore, $2\sin^{-1}\left(\frac12\right)=2\cdot \frac{\pi}{6}=\frac{\pi}{3}.$

Match with options

$\boxed{\frac{\pi}{3}}$

So the correct option is B.